using Functor for map values?

[tomerdr]

Hi,
I have the following function and functor and i want to print all
values of a map
but pair does not support operator << of course.
so i wonder what are my options?
can i extend the pair struct? can i create a values iterator?
how do i do it?

Thanks in advance.

template <class Iterator, class Function>
void apply(Iterator begin, Iterator end, const Function& f)
{
for(;begin != end; ++begin)
f(*begin);
}

class Print
{
public:
template<class Tvoid operator()(cons t T& printable) const
{
cout << printable << endl;
}
};

[=?iso-8859-1?q?Erik_Wikstr=F6m?=]

On Feb 1, 10:38 am, tome...@gmail.c om wrote:

Hi,
I have the following function and functor and i want to print all
values of a map
but pair does not support operator << of course.
so i wonder what are my options?
can i extend the pair struct? can i create a values iterator?
how do i do it?

Thanks in advance.

template <class Iterator, class Function>
void apply(Iterator begin, Iterator end, const Function& f)
{
for(;begin != end; ++begin)
f(*begin);

}

class Print
{
public:
template<class Tvoid operator()(cons t T& printable) const
{
cout << printable << endl;
}

};

Do you want to print the key-value pair of just the value? If you want
the second you could change the line 'f(*begin);' to 'f(begin-

>second);', this would however make the function a lot less general. A

better approach, which works regardless if you want to print just the
value or the key-value pair, is to create an overloaded version of the
()-operator.

class Print
{
public:
template<class T>
void operator()(cons t T& printable) const
{
cout << printable << endl;
}
template<class T, class U>
void operator()(cons t pair<T, U>& printable) const
{
cout << printable.first << ": " << printable.secon d << endl;
}
};

--
Erik Wikström

[tomerdr]

I want to print the value only and it must be a general functor.

Thanks.

[=?iso-8859-1?q?Erik_Wikstr=F6m?=]

On Feb 1, 11:38 am, tome...@gmail.c om wrote:

I want to print the value only and it must be a general functor.

First of all, try to quote relevant sections of the post you are
replying to.

No, either you can take the code I posted in the previous message and
modify it a bit so that it only prints the value. Or you can define a
<<-operator for std::pair:

template<class T, class U>
std::ostream& operator<<(std: :ostream& os, const std::pair<T, U>& p)
{
os << p.second;
return os;
}

--
Erik Wikström

[dasjotre]

Or you can define a

<<-operator for std::pair:

template<class T, class U>
std::ostream& operator<<(std: :ostream& os, const std::pair<T, U>& p)
{
os << p.second;
return os;

}

Wouldn't that fail due to ADL unless it is put into std namespace?

[=?iso-8859-1?q?Erik_Wikstr=F6m?=]

On Feb 2, 4:11 pm, "dasjotre" <dasjo...@googl email.comwrote:

Or you can define a
<<-operator for std::pair:

template<class T, class U>
std::ostream& operator<<(std: :ostream& os, const std::pair<T, U>& p)
{
os << p.second;
return os;

}

Wouldn't that fail due to ADL unless it is put into std namespace?

Hmm, don't know. It worked in my little test program but that was just
one small file with all the code in it, so there might be cases where
it won't work.

--
Erik Wikström

[Dave Rahardja]

On 2 Feb 2007 07:11:45 -0800, "dasjotre" <da******@googl email.comwrote:

>Or you can define a
<<-operator for std::pair:

template<cla ss T, class U>
std::ostream & operator<<(std: :ostream& os, const std::pair<T, U>& p)
{
os << p.second;
return os;

}

Wouldn't that fail due to ADL unless it is put into std namespace?

No. ADL will _include_ any operator<<(std: :ostream&, const std::pair<T,U>& ) in
std in the list of candidate functions, but it will not _exclude_ the version
that is defined in your current scope.

-dr